Problem: Simplify and expand the following expression: $ \dfrac{4}{3z - 12}- \dfrac{2}{z - 3}+ \dfrac{2z}{z^2 - 7z + 12} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3z - 12} = \dfrac{4}{3(z - 4)}$ We can factor the quadratic in the third term: $ \dfrac{2z}{z^2 - 7z + 12} = \dfrac{2z}{(z - 4)(z - 3)}$ Now we have: $ \dfrac{4}{3(z - 4)}- \dfrac{2}{z - 3}+ \dfrac{2z}{(z - 4)(z - 3)} $ The least common multiple of the denominators is: $ 3(z - 4)(z - 3)$ In order to get the first term over $3(z - 4)(z - 3)$ , multiply by $\dfrac{z - 3}{z - 3}$ $ \dfrac{4}{3(z - 4)} \times \dfrac{z - 3}{z - 3} = \dfrac{4(z - 3)}{3(z - 4)(z - 3)} $ In order to get the second term over $3(z - 4)(z - 3)$ , multiply by $\dfrac{3(z - 4)}{3(z - 4)}$ $ \dfrac{2}{z - 3} \times \dfrac{3(z - 4)}{3(z - 4)} = \dfrac{6(z - 4)}{3(z - 4)(z - 3)} $ In order to get the third term over $3(z - 4)(z - 3)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{2z}{(z - 4)(z - 3)} \times \dfrac{3}{3} = \dfrac{6z}{3(z - 4)(z - 3)} $ Now we have: $ \dfrac{4(z - 3)}{3(z - 4)(z - 3)} - \dfrac{6(z - 4)}{3(z - 4)(z - 3)} + \dfrac{6z}{3(z - 4)(z - 3)} $ $ = \dfrac{ 4(z - 3) - 6(z - 4) + 6z} {3(z - 4)(z - 3)} $ Expand: $ = \dfrac{4z - 12 - 6z + 24 + 6z}{3z^2 - 21z + 36} $ $ = \dfrac{4z + 12}{3z^2 - 21z + 36}$